40(x+3)-3x(x+3)=40x — step-by-step solution

Let's solve in

ℝ

40 \cdot (x + 3) - 3 x \cdot (x + 3) = 40 x

40 \cdot (x + 3) - 3 x \cdot (x + 3) - 40 x = 0

ⓘ Expand by multiplying each term inside the parentheses:

40 \cdot (x + 3)

40 x + 120

- 3 x \cdot (x + 3)

- (3 x^{2} + 9 x)

40 x

+ 120

- (3 x^{2} + 9 x)

- 40 x = 0

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (3 x^{2} + 9 x)

- 3 x^{2} - 9 x

■Like terms are grouped by adding their coefficients:

40 x - 40 x

0

0

+ 120

- 3 x^{2}

- 9 x = 0

ⓘ Adding 0 changes nothing:

0 + 120

120

120 - 3 x^{2} - 9 x = 0

Since

-3 \neq 0

, divide both sides by

-3

- 40 + x^{2} + 3 x = 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 3

and

c = -40

Δ = 3^{2} - 4 \cdot 1 \cdot (-40)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

3^{2}

3 \cdot 3

9

Δ = 9 - 4 \cdot (-40)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-40)

160

Δ = 9 + 160

ⓘ Start from

9

and move forward

160

on the number line:

9 + 160

169

Δ = 169

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 3 - \sqrt{169}}{2 \cdot 1}

x_{1} = \frac{- 3 - \sqrt{169}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{169}

13

;

13^{2}

169

x_{1} = \frac{- 3 - 13}{2}

ⓘ The numerator and the denominator are computed before the division:

- 3 - 13

-16

x_{1} = - \frac{16}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{16}{2}

-8

x_{1} = -8

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 3 + \sqrt{169}}{2 \cdot 1}

x_{2} = \frac{- 3 + \sqrt{169}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{169}

13

;

13^{2}

169

x_{2} = \frac{- 3 + 13}{2}

ⓘ The numerator and the denominator are computed before the division:

- 3 + 13

10

x_{2} = \frac{10}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{10}{2}

5

x_{2} = 5

▶Solutions found:

x = {-8, 5}